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This is my first time asking something around here, so I hope I formulate the question in the best way possible.

I've been learning python for a couple of months now and I decided to put it to use for a problem I have to solve. It involves some deal of linear algebra, so I'll try to establish the maths first. Basically, suppose that you are given a vector in some basis (to be concise, I'll write everything in two dimensions) which we write as

v = (a*e_1 + b*e_2, c*e_1 + d*e_2)

which means that I can get v as the product of the vector (e_1,e_2) and a 2x2 matrix ((a,b),(c,d)). Now, my problem is the following: given the vector v, I want to extract the matrix ((a,b),(c,d)) and put it into a numpy array. Since eventually I will need to do this for a list of vectors, I would like to be able to give as input a list with entries in the form of strings which represents the vector, say:

vector = [2*e_1, -e_1 + e_2]

and get as output the matrix, as a numpy array, that projects "vector" onto the basis, that is

matrix = np.array([[2, 0], [-1, 2]])

is this possible? And, if so, can anyone point me in the right direction? I have been staring at the screen for a bit and haven't figured out how to do it, so any kind of hint would be very much appreciated.

EDIT:

So, thanks to everyone that has answered so far. Let me expand upon the problem I have and what I want to do with it.

I have, as input, a set of vectors whose components are all linear combinations of a set of basis vectors (k1,k2,...,kn) with coefficients in the set (+1,-1,0). This set of vectors is of size N, where in general, N >= n. Then, given a choice of n vectors out of the N input vectors, I can regards those n vectors, which I'll call (k1',k2',...,kn') as a new basis and compute the matrix which transforms from the basis (k1,k2,...,kn) to the new basis (k1',k2',...,kn'). Moreover, I can do this with a second choice of another n vectors and get the transformation matrix to a third basis (k1'',k2'',...,kn'').

What my problem really needs is for me to find the matrix that transforms from the basis b1 = (k1',k2',...,kn') to the basis b2 = (k1'',k2'',...,kn''), given that the input is written in the basis b = (k1,k2,...,kn). However, I know that if I write b1 = Ab and b2 = Bb for some matrices A and B, I can find the matrix C in the relation b1 = Cb2 by finding the inverse for B and writting C=A*B^{-1}. After posting this question and playing with stuff, I came up with the following code:

    import numpy as np
    import itertools

    input_vectors = [['+l1', '+l2'], ['+l1+l2', '+l1']]
    dim = len(input_vectors)
    matrix_list = [np.zeros((dim, dim)) for i in input_vectors]

    for element in zip(input_vectors, matrix_list):
        momenta = element[0]
        matrix = element[1]
        if '+l1' in momenta[0]:
            matrix[0][0] = 1
        if '+l2' in momenta[0]:
            matrix[0][1] = 1
        if '-l1' in momenta[0]:
            matrix[0][0] = -1
        if '-l2' in momenta[0]:
            matrix[0][1] = -1

        if '+l1' in momenta[1]:
            matrix[1][0] = 1
        if '+l2' in momenta[1]:
            matrix[1][1] = 1
        if '-l1' in momenta[1]:
            matrix[1][0] = -1
        if '-l2' in momenta[1]:
            matrix[1][1] = -1

which actually yields my desired output. However, it looks awfully redundant and evidently not easy to generalize.

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  • How are you currently representing vector = [2*e_1, -e_1 + e_2] in python? What are the data types of e_1 and e_2? Commented Apr 15, 2021 at 20:03
  • Currently, I'm thinking of representing e_1 and e_2 as strings (the actual coefficients of e_1 and e_2 in the problem are always going to equal 1, -1 or 0, but I need something that makes a distinction between e_1 and e_2). However, I don't even know if that is the best approach. Commented Apr 15, 2021 at 20:15
  • Wow, I just got confused with your example np.array([[2, 0], [-1, 2]]), is it really 2 in the last position? Is your matrix transposed? Commented Apr 15, 2021 at 20:23

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What are you trying to do with this linear algebra? So you want to have this [2*e_1, -e_1 + e_2] as input and this np.array([[2, 0], [-1, 2]]) as output?

Parsing the string is just too painful, so let's supposed you just wrote that expression in Python (if not use exec):

string_expression = "[2*e_1, -e_1 + e_2]" # If you read strings from file or input
e_1 = np.array([1, 0]) # Those vectors need to be defined before eval
e_2 = np.array([0, 1])
python_expression = eval(string_expression) # You don't need this if you just write the python expression yourself, like so
# python_expression = [2*e_1, -e_1 + e_2] # This is simpler
matrix = np.array(python_expression)

If you want a more generic approach:

def get_matrix(string_expression, dims, vector_name='e'):
     for i in range(dims):
          base = np.zeros(dims)
          base[i] = 1 # It just a vector in the form [0, 0, ... ,1, ... 0, 0]
          base_name = '%s_%d' % (vector_name, i+1) # string for e_i
          locals()[base_name] = base # create a local variable named e_i
     expression = eval(string_expression) # evaluates the string as Python code
     return np.array(expression) # Create a matrix from the list of arrays

get_matrix('[2*e_1, -e_1 + e_2]', 2, 'e') # returns np.array([[2, 0], [-1, 1]])
get_matrix('[l_1+l2, l_1]', 2, 'l') # returns np.array([[1, 1], [1, 0]])
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3 Comments

Thanks a lot for your answer! I just made an edit to the original question, which hopefully helps making my problem and the motivation for it clearer. I have one question, though: what does the .T in matrix = np.array(python_expression).T do? I have only done vary basic stuff with numpy so I don't know a lot about the syntax and methods
It is to transpose the matrix, .T is a short for np.transpose. Doesn't this answer work? I have also made a generic function, it accepts any python expression. You should know that the identity values for the vectors (one-hot encoding) are what creates the matrix.
Hi! Thanks for the answer! I just came back to this, and it definitely does the job. I didn't completely understand the logic before, so I was curious about that.

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